Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 1}{x - 4} = \dfrac{3x + 5}{x - 4}$
Explanation: Multiply both sides by $x - 4$ $ \dfrac{x^2 + 1}{x - 4} (x - 4) = \dfrac{3x + 5}{x - 4} (x - 4)$ $ x^2 + 1 = 3x + 5$ Subtract $3x + 5$ from both sides: $ x^2 + 1 - (3x + 5) = 3x + 5 - (3x + 5)$ $ x^2 + 1 - 3x - 5 = 0$ $ x^2 - 4 - 3x = 0$ Factor the expression: $ (x - 4)(x + 1) = 0$ Therefore $x = 4$ or $x = -1$ At $x = 4$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 4$, it is an extraneous solution.